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diff --git a/2-point.tex b/2-point.tex index bb1b625..380d677 100644 --- a/2-point.tex +++ b/2-point.tex @@ -37,6 +37,8 @@ \cite{Ros_2020_Distribution, Ros_2019_Complex, Ros_2019_Complexity} +\section{Model} + The mixed $p$-spin models are defined by the Hamiltonian \begin{equation} \label{eq:hamiltonian} H(\mathbf s)=-\sum_p\frac1{p!}\sum_{i_1\cdots i_p}^NJ^{(p)}_{i_1\cdots i_p}s_{i_1}\cdots s_{i_p} @@ -48,7 +50,7 @@ the energy is typically extensive. The overbar denotes an average over the coefficients $J$. The factors $a_p$ in the variances are freely chosen constants that define the particular model. For instance, the `pure' $p$-spin model has $a_{p'}=\delta_{p'p}$. This class of models encompasses all -statistically isotropic gaussian random Hamiltonians defined on the +statistically isotropic Gaussian random Hamiltonians defined on the hypersphere. The covariance between the energy at two different points is a function of the overlap, or dot product, between those points, or @@ -59,52 +61,92 @@ where the function $f$ is defined from the coefficients $a_p$ by \begin{equation} f(q)=\frac12\sum_pa_pq^p \end{equation} -In this paper, we will focus on models with a replica symmetric complexity. +In this paper, we will focus on models with a replica symmetric complexity, but +many of the intermediate formulae are valid for arbitrary replica symmetry +breakings. + +To enforce the spherical constraint at stationary points, we make use of a Lagrange multiplier $\omega$. This results in the extremal problem +\begin{equation} + H(\mathbf s)+\frac\omega2(\|\mathbf s\|^2-N) +\end{equation} +The gradient and Hessian at a stationary point are then +\begin{align} + \nabla H(\mathbf s,\omega)=\partial H(\mathbf s)+\omega\mathbf s + && + \operatorname{Hess}H(\mathbf s,\omega)=\partial\partial H(\mathbf s)+\omega I +\end{align} +where $\partial=\frac\partial{\partial\mathbf s}$ will always denote the derivative with respect to $\mathbf s$. We introduce the Kac--Rice \cite{Kac_1943_On, Rice_1944_Mathematical} measure \begin{equation} - d\nu_H(\mathbf s) - =d\mathbf s\,\delta\big(\nabla H(\mathbf s)\big)\, - \big|\det\operatorname{Hess}H(\mathbf s)\big| + d\nu_H(\mathbf s,\omega) + =2\,d\mathbf s\,d\omega\,\delta(\|\mathbf s\|^2-N)\, + \delta\big(\nabla H(\mathbf s,\omega)\big)\, + \big|\det\operatorname{Hess}H(\mathbf s,\omega)\big| \end{equation} which counts stationary points of the function $H$. More interesting is the measure conditioned on the energy density $E$ and stability $\mu$, \begin{equation} - d\nu_H(\mathbf s\mid E,\mu) - =d\nu_H(\mathbf s)\, + d\nu_H(\mathbf s,\omega\mid E,\mu) + =d\nu_H(\mathbf s,\omega)\, \delta\big(H(\mathbf s)-NE\big)\, - \delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf s)\big) + \delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf s,\omega)\big) \end{equation} +\section{Complexity} We want the typical number of stationary points with energy density -$E_2$ and stability $\mu_2$ that lie a fixed overlap $q$ from a reference -stationary point of energy density $E_1$ and stability $\mu_1$. -\begin{align*} +$E_1$ and stability $\mu_1$ that lie a fixed overlap $q$ from a reference +stationary point of energy density $E_0$ and stability $\mu_0$. +\begin{equation} \label{eq:complexity.definition} \Sigma_{12} - &=\frac1N\overline{\int\frac{d\nu_H(\mathbf s_0\mid E_0,\mu_0)}{\int d\nu_H(\mathbf s_0'\mid E_0,\mu_0)}\, - \log\bigg(\int d\nu_H(\mathbf s_1\mid E_1,\mu_1)\,\delta(Nq-\mathbf s_0\cdot\mathbf s_1)\bigg)} -\end{align*} -\begin{align*} + =\frac1N\overline{\int\frac{d\nu_H(\pmb\sigma,\varsigma\mid E_0,\mu_0)}{\int d\nu_H(\pmb\sigma',\varsigma'\mid E_0,\mu_0)}\, + \log\bigg(\int d\nu_H(\mathbf s,\omega\mid E_1,\mu_1)\,\delta(Nq-\pmb\sigma\cdot\mathbf s)\bigg)} +\end{equation} +Both the denominator and the logarithm are treated using the replica trick, which yields +\begin{equation} \Sigma_{12} - &=\frac1N\lim_{n\to0}\lim_{m\to-1}\overline{\int d\nu_H(\mathbf s_0\mid E_0,\mu_0)\left(\int d\nu_H(\mathbf s_0'\mid E_0,\mu_0)\right)^m\, - \frac\partial{\partial n}\bigg(\int d\nu_H(\mathbf s_1\mid E_1,\mu_1)\,\delta(Nq-\mathbf s_0\cdot \mathbf s_1)\bigg)^n}\\ - &=\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\overline{\int\left(\prod_{b=1}^md\nu_H(\pmb\sigma_b\mid E_0,\mu_0)\right)\left(\prod_{a=1}^nd\nu_H(\mathbf s_a\mid E_1,\mu_1)\,\delta(Nq-\pmb \sigma_1\cdot \mathbf s_a)\right)} -\end{align*} + =\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\overline{\int\left(\prod_{b=1}^md\nu_H(\pmb\sigma_b,\varsigma_b\mid E_0,\mu_0)\right)\left(\prod_{a=1}^nd\nu_H(\mathbf s_a,\omega_a\mid E_1,\mu_1)\,\delta(Nq-\pmb \sigma_1\cdot \mathbf s_a)\right)} +\end{equation} +Note that because of the structure of \eqref{eq:complexity.definition}, +$\pmb\sigma_1$ is special among the set of $\pmb\sigma$ replicas, since only it +is constrained to lie a given overlap from the $\mathbf s$ replicas. This +replica asymmetry will be important later. + +\subsection{The Hessian factors} + +The double partial derivatives of the energy are Gaussian with the variance \begin{equation} - \overline{\big|\det\operatorname{Hess}H(s)\big|\,\delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(s)\big)} - =e^{N\int d\lambda\,\rho(\lambda+\mu)\log|\lambda|}\delta(N\mu-s\cdot\partial H) + \overline{(\partial_i\partial_jH(\mathbf s))^2}=\frac1Nf''(1) \end{equation} +which means that the matrix of partial derivatives belongs to the GOE class. Its spectrum is given by the Wigner semicircle \begin{equation} \rho(\lambda)=\begin{cases} \frac2{\pi}\sqrt{1-\big(\frac{\lambda}{\mu_\text m}\big)^2} & \lambda^2\leq\mu_\text m^2 \\ 0 & \text{otherwise} \end{cases} \end{equation} +with radius $\mu_\text m=\sqrt{4f''(1)}$. Since the Hessian differs from the +matrix of partial derivatives by adding the constant diagonal matrix $\omega +I$, it follows that the spectrum of the Hessian is a Winger semicircle shifted +by $\omega$, or $\rho(\lambda+\omega)$. + +The average over factors depending on the Hessian alone can be made separately +from those depending on the gradient or energy, since for random Gaussian +fields the Hessian is independent of these \cite{Bray_2007_Statistics}. In +principle the fact that we have conditioned the Hessian to belong to stationary +points of certain energy, stability, and proximity to another stationary point +will modify its statistics, but these changes will only appear at subleading +order in $N$ \cite{Ros_2019_Complexity}. At leading order, the various expectations factorize, each yielding +\begin{equation} + \overline{\big|\det\operatorname{Hess}H(\mathbf s,\omega)\big|\,\delta\big(N\mu-\operatorname{Tr}\operatorname{Hess}H(\mathbf s,\omega)\big)} + =e^{N\int d\lambda\,\rho(\lambda+\mu)\log|\lambda|}\delta(N\mu-N\omega) +\end{equation} +Therefore, all of the Lagrange multipliers are fixed identically to the stabilities $\mu$. We define the function \begin{equation} \begin{aligned} \mathcal D(\mu) - &=\int d\lambda\,\rho(\lambda+\mu)\ln|\lambda| \\ + &=\int d\lambda\,\rho(\lambda+\mu)\log|\lambda| \\ &=\begin{cases} \frac12+\log\left(\frac12\mu_\text m\right)+\frac\mu{\mu_\text m}\left(\frac\mu{\mu_\text m}-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) -\log\left(\frac{\mu}{\mu_\text m}-\sqrt{\big(\frac\mu{\mu_\text m}\big)^2-1}\right) & \mu>\mu_\text m \\ @@ -115,28 +157,92 @@ stationary point of energy density $E_1$ and stability $\mu_1$. \end{cases} \end{aligned} \end{equation} +and the full factor due to the Hessians is +\begin{equation} + e^{Nm\mathcal D(\mu_0)+Nn\mathcal D(\mu_1)}\left[\prod_a^m\delta(N\mu_0-N\varsigma_a)\right]\left[\prod_a^n\delta(N\mu_1-N\omega_a)\right] +\end{equation} -\begin{align} - \mathcal Q_{00}=\begin{bmatrix} - \hat\beta_0\\\hat\mu_0\\C^{00}\\R^{00}\\D^{00} - \end{bmatrix} - && - \mathcal Q_{11}=\begin{bmatrix} - \hat\beta_1\\\hat\mu_1\\C^{11}\\R^{11}\\D^{11} - \end{bmatrix} - && - \mathcal Q_{01}=\begin{bmatrix} - \hat\mu_{01}\\C^{01}\\R^{01}\\R_{10}\\D^{01} - \end{bmatrix} +\subsection{The other factors} + +Having integrated over the Lagrange multipliers using the $\delta$ functions +resulting from the average of the Hessians, the remaining part of the integrand +has the form +\begin{equation} + e^{ + -Nm\hat\beta_0E_0-Nn\hat\beta_1E_1 + -\sum_a^m\left[(\pmb\sigma_a\cdot\hat{\pmb\sigma}_a)\mu_0 + -\frac12\hat\mu_0(N-\pmb\sigma_a\cdot\pmb\sigma_a) + \right] + -\sum_a^n\left[(\mathbf s_a\cdot\hat{\mathbf s}_a)\mu_1 + -\frac12\hat\mu_1(N-\mathbf s_a\cdot\mathbf s_a) + -\frac12\hat\mu_{12}(Nq-\pmb\sigma_1\cdot\mathbf s_a) + \right] + +\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t) + } +\end{equation} +where we have introduced the linear operator +\begin{equation} + \mathcal O(\mathbf t) + =\sum_a^m\delta(\mathbf t-\pmb\sigma_a)\left( + i\hat{\pmb\sigma}_a\cdot\partial_{\mathbf t}+\hat\beta_0 + \right) + + + \sum_a^n\delta(\mathbf t-\mathbf s_a)\left( + i\hat{\mathbf s}_a\cdot\partial_{\mathbf t}+\hat\beta_1 + \right) +\end{equation} +We have written the $H$-dependant terms in this strange form for the ease of taking the average over $H$: since it is Gaussian-correlated, it follows that +\begin{equation} + \overline{e^{\int d\mathbf t\,\mathcal O(\mathbf t)H(\mathbf t)}} + =e^{\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')\overline{H(\mathbf t)H(\mathbf t')}} + =e^{N\frac12\int d\mathbf t\,d\mathbf t'\,\mathcal O(\mathbf t)\mathcal O(\mathbf t')f\big(\frac{\mathbf t\cdot\mathbf t'}N\big)} +\end{equation} +It remains only to apply the doubled operators to $f$ and then evaluate the simple integrals over the $\delta$ measures. We do not include these details, which are standard. + +\subsection{Hubbard--Stratonovich} + +Having expanded this expression, we are left with an argument in the exponential which is a function of scalar products between the fields $\mathbf s$, $\hat{\mathbf s}$, $\pmb\sigma$, and $\hat{\pmb\sigma}$. We will change integration coordinates from these fields to matrix fields given by the scalar products, defined as +\begin{align} \label{eq:fields} + C^{00}_{ab}=\frac1N\pmb\sigma_a\cdot\pmb\sigma_b && + R^{00}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b && + D^{00}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\pmb\sigma}_b \\ + C^{01}_{ab}=\frac1N\pmb\sigma_a\cdot\mathbf s_b && + R^{01}_{ab}=-i\frac1N{\pmb\sigma}_a\cdot\hat{\mathbf s}_b && + R^{10}_{ab}=-i\frac1N\hat{\pmb\sigma}_a\cdot{\mathbf s}_b && + D^{01}_{ab}=\frac1N\hat{\pmb\sigma}_a\cdot\hat{\mathbf s}_b \\ + C^{11}_{ab}=\frac1N\mathbf s_a\cdot\mathbf s_b && + R^{11}_{ab}=-i\frac1N{\mathbf s}_a\cdot\hat{\mathbf s}_b && + D^{11}_{ab}=\frac1N\hat{\mathbf s}_a\cdot\hat{\mathbf s}_b \end{align} +We insert into the integral the product of $\delta$ functions enforcing these +definitions, integrated over the new matrix fields, which is equivalent to +multiplying by one. Once this is done, the many scalar products appearing +throughout can be replaced by the matrix fields, and the original vector fields +can be integrated over. Conjugate matrix field integrals created when the +$\delta$ functions are promoted to exponentials can be evaluated by saddle +point in the standard way, yielding an effective action depending on the above +matrix fields alone. + +We will always assume that the square matrices $C^{00}$, $R^{00}$, $D^{00}$, +$C^{11}$, $R^{11}$, and $D^{11}$ are hierarchical matrices, with each set of +three sharing the same hierarchical structure. In particular, we immediately +define $c_\mathrm d^{00}$, $r_\mathrm d^{00}$, $d_\mathrm d^{00}$, $c_\mathrm d^{11}$, $r_\mathrm d^{11}$, and +$d_\mathrm d^{11}$ as the value of the diagonal elements of these matrices, +respectively. Note that $c_\mathrm d^{00}=c_\mathrm d^{11}=1$ due to the spherical constraint. + +Defining the `block' fields $\mathcal Q_{00}=(\hat\beta_0, \hat\mu_0, C^{00}, +R^{00}, D^{00})$, $\mathcal Q_{11}=(\hat\beta_1, \hat\mu_1, C^{11}, R^{11}, +D^{11})$, and $\mathcal Q_{01}=(\hat\mu_{01},C^{01},R^{01},R^{10},D^{01})$ +the resulting complexity is \begin{equation} \Sigma_{01} =\frac1N\lim_{n\to0}\lim_{m\to0}\frac\partial{\partial n}\int d\mathcal Q_{00}\,d\mathcal Q_{11}\,d\mathcal Q_{01}\,e^{Nm\mathcal S_0(\mathcal Q_{00})+Nn\mathcal S_1(\mathcal Q_{00},\mathcal Q_{11},\mathcal Q_{01})} \end{equation} +where \begin{equation} \begin{aligned} &\mathcal S_0(\mathcal Q_{00}) - =-\hat\beta_0E_0-r^{00}_d\mu_0-\frac12\hat\mu_0(1-c^{00}_d)+\mathcal D(\mu_0)\\ + =-\hat\beta_0E_0-r^{00}_\mathrm d\mu_0-\frac12\hat\mu_0(1-c^{00}_\mathrm d)+\mathcal D(\mu_0)\\ &\quad+\frac1m\bigg\{ \frac12\sum_{ab}^m\left[ \hat\beta_1^2f(C^{00}_{ab})-(2\hat\beta_1R^{00}_{ab}+D^{00}_{ab})f'(C^{00}_{ab})+(R_{ab}^{00})^2f''(C_{ab}^{00}) @@ -144,11 +250,11 @@ stationary point of energy density $E_1$ and stability $\mu_1$. \bigg\} \end{aligned} \end{equation} - +is the action for the ordinary, one-point complexity, and remainder is given by \begin{equation} \begin{aligned} &\mathcal S(\mathcal Q_{00},\mathcal Q_{11},\mathcal Q_{01}) - =-\hat\beta_1E_1-\mu_1r^{11}_d-\frac12\hat\mu_1(1-c^{11}_d) \\ + =-\hat\beta_1E_1-r^{11}_\mathrm d\mu_1-\frac12\hat\mu_1(1-c^{11}_\mathrm d)+\mathcal D(\mu_1) \\ &\quad+\frac1n\sum_b^n\left\{-\frac12\hat\mu_{12}(q-C^{01}_{1b})+\sum_a^m\left[ \hat\beta_0\hat\beta_1f(C^{01}_{ab})-(\hat\beta_0R^{01}_{ab}+\hat\beta_1R^{10}_{ab}+D^{01}_{ab})f'(C^{01}_{ab})+R^{01}_{ab}R^{10}_{ab}f''(C^{01}_{ab}) \right]\right\} @@ -174,10 +280,15 @@ stationary point of energy density $E_1$ and stability $\mu_1$. \bigg\} \end{aligned} \end{equation} +Because of the structure of this problem in the twin limits of $m$ and $n$ to +zero, the parameters $\mathcal Q_{00}$ can be evaluated at a saddle point of +$\mathcal S_0$ alone. This means that these parameters will take the same value +they take when the ordinary, 1-point complexity is calculated. +The $m\times n$ matrices $C^{01}$, $R^{01}$, $R^{10}$, and $D^{01}$ are +expected to have the following form at the saddle point: \begin{align} - C^{01} - = + C^{01}= \begin{subarray}{l} \hphantom{[}\begin{array}{ccc}\leftarrow&n&\rightarrow\end{array}\hphantom{\Bigg]}\\ \left[ @@ -189,8 +300,7 @@ stationary point of energy density $E_1$ and stability $\mu_1$. \end{array} \right]\begin{array}{c} \\\uparrow\\m-1\\\downarrow - \end{array}\\ - \vphantom{\begin{array}{c}n\end{array}} + \end{array} \end{subarray} && R^{01} @@ -217,8 +327,10 @@ stationary point of energy density $E_1$ and stability $\mu_1$. 0&\cdots&0 \end{bmatrix} \end{align} +where only the first row is nonzero as a result of the sole linear term +proportional to $C_{1b}^{01}$ in the action. -The inverse of block hierarchical matrix is still a block hierarchical matrix, since (dropping the superscripts for clarity) +The inverse of block hierarchical matrix is still a block hierarchical matrix, since \begin{equation} \begin{bmatrix} C^{00}&iR^{00}\\iR^{00}&D^{00} @@ -231,72 +343,35 @@ The inverse of block hierarchical matrix is still a block hierarchical matrix, s \end{equation} Because of the structure of the 01 matrices, the volume element will depend only on the diagonal if this matrix. If we write \begin{align} - \tilde c_d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}]_{11} \\ - \tilde r_d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00}]_{11} \\ - \tilde d_d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00}]_{11} + \tilde c_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}C^{00}]_{\text d} \\ + \tilde r_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}R^{00}]_{\text d} \\ + \tilde d_\mathrm d^{00}&=[(C^{00}D^{00}+R^{00}R^{00})^{-1}D^{00}]_{\text d} \end{align} - -In the replica symmetric case, -\begin{align} - \tilde c_d^{00}=\frac1{(r^{00}_d)^2+d^{00}_d} && - \tilde r_d^{00}=\frac{r^{00}_d}{(r^{00}_d)^2+d^{00}_d} && - \tilde d_d^{00}=\frac{d^{00}_d}{(r^{00}_d)^2+d^{00}_d} -\end{align} - +then the result is \begin{equation} - \begin{bmatrix} - q^2\tilde d_d^{00}+2qr_{10}\tilde r^{00}_d-r_{10}^2\tilde d^{00}_d - & - i\left[d_{01}(r_{10}\tilde c^{00}_d-q\tilde r^{00}_d)+r_{01}(r_{10}\tilde r^{00}_d+q\tilde d^{00}_d)\right] - \\ - i\left[d_{01}(r_{10}\tilde c^{00}_d-q\tilde r^{00}_d)+r_{01}(r_{10}\tilde r^{00}_d+q\tilde d^{00}_d)\right] - & - d_{01}^2\tilde c^{00}_d+2r_{01}d_{01}\tilde r^{00}_d-r_{01}^2\tilde d^{00}_d - \end{bmatrix} + \begin{aligned} + & \begin{bmatrix} + C^{01}&iR^{01}\\iR^{10}&D^{01} + \end{bmatrix}^T + \begin{bmatrix} + C^{00}&iR^{00}\\iR^{00}&D^{00} + \end{bmatrix}^{-1} + \begin{bmatrix} + C^{01}&iR^{01}\\iR^{10}&D^{01} + \end{bmatrix} \\ + &\qquad=\begin{bmatrix} + q^2\tilde d_\mathrm d^{00}+2qr_{10}\tilde r^{00}_\mathrm d-r_{10}^2\tilde d^{00}_\mathrm d + & + i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right] + \\ + i\left[d_{01}(r_{10}\tilde c^{00}_\mathrm d-q\tilde r^{00}_\mathrm d)+r_{01}(r_{10}\tilde r^{00}_\mathrm d+q\tilde d^{00}_\mathrm d)\right] + & + d_{01}^2\tilde c^{00}_\mathrm d+2r_{01}d_{01}\tilde r^{00}_\mathrm d-r_{01}^2\tilde d^{00}_\mathrm d + \end{bmatrix} + \end{aligned} \end{equation} where each block is a constant $n\times n$ matrix. -In the twin limits of $m$ and $n$ to zero, the saddle point conditions for the variables involving only the reference critical point (those in $\mathcal Q_{00}$) reduce to the ordinary, 1-point conditions. With a replica-symmetric ansatz, these conditions are -\begin{align} - \hat\beta_0 - &=-\frac{(\epsilon_0+\mu_0)f'(1)+\epsilon_0f''(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}\\ - r_d^{00} - &=\frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} \\ - d_d^{00} - &=\frac1{f'(1)} - -\left( - \frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} - \right)^2 -\end{align} - -\begin{align*} - & - \begin{bmatrix} - \tilde c&\tilde r\\\tilde r&\tilde d - \end{bmatrix} - = - \begin{bmatrix} - q&ir_{10}\\ir_{01}&d_{01} - \end{bmatrix} - \begin{bmatrix} - 1&ir_{0}\\ - ir_{0}&d_{0} - \end{bmatrix}^{-1} - \begin{bmatrix} - q&ir_{01}\\ir_{10}&d_{01} - \end{bmatrix}\\ - &= - \frac1{r_{0}^2+d_{0}}\begin{bmatrix} - q^2d_{0}+2qr_{0}r_{10}-r_{10}^2 - & - i\left[d_{01}(r_{10}-r_0q)+r_{01}(r_0r_{10}+d_0q)\right] - \\ - i\left[d_{01}(r_{10}-r_0q)+r_{01}(r_0r_{10}+d_0q)\right] - & - d_{01}^2+2r_{0}r_{01}d_{01}-d_{0}r_{01}^2 - \end{bmatrix} -\end{align*} -This matrix with modify the diagonal of the RS matrix for the second spin. Define $\tilde C=C-\tilde c$, $\tilde R=R-\tilde r$, $\tilde D=D-\tilde d$. Then \begin{align*} @@ -351,6 +426,29 @@ Solving, we get \right) \end{align*} +\section{Replica symmetric case} + +With a replica-symmetric ansatz, these conditions are +\begin{align} + \hat\beta_0 + &=\frac{(\epsilon_0+\mu_0)f'(1)+\epsilon_0f''(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2}\\ + r_\mathrm d^{00} + &=\frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} \\ + d_\mathrm d^{00} + &=\frac1{f'(1)} + -\left( + \frac{\mu_0f(1)+\epsilon_0f'(1)}{f(1)\big(f'(1)+f''(1)\big)-f'(1)^2} + \right)^2 +\end{align} + +In the replica symmetric case, +\begin{align} + \tilde c_\mathrm d^{00}=\frac1{(r^{00}_\mathrm d)^2+d^{00}_\mathrm d} && + \tilde r_\mathrm d^{00}=\frac{r^{00}_\mathrm d}{(r^{00}_\mathrm d)^2+d^{00}_\mathrm d} && + \tilde d_\mathrm d^{00}=\frac{d^{00}_\mathrm d}{(r^{00}_\mathrm d)^2+d^{00}_\mathrm d} +\end{align} + + \begin{equation} \hat\beta_2E_2-r_{22}^{(0)}\mu_2\frac12\left\{ \hat\beta_2^2\big(f(1)-f(q_{22}^{(0)})\big) |